Comparing Olympus 4/3lenses to FX "Full Frame" offerings

Started Jan 25, 2014 | Discussions thread
bobn2 Forum Pro • Posts: 58,518
Re: Wrong

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know. That is what I am saying! The intensity of light per unit area is the same...... I thought that I have made that obvious.

Then why would you say just above “But that light is spread over 4 times the area “ when responding to uncle Bobs “ thus a ff collect’s 4 times the light at the same F number”

Don’t you see the fallacy of say” but that light is spread over 4 time the area” when the intensity remains the same ?

Uncle Bob is just plain wrong.

No, Uncle Bob is right. Think what 'per unit area' means. For each similar unit of area, the two sensors receive the same light. Let's choose the unit to be a square millimetre. The Four Thirds sensor is 225 square mm, the FF one is 864, nearly four times (actually 3.84). So if they both receive the same light per square mm, then the FF sensor receives 3.84 times the light.

He claims that each pixel receives 4 times as much light. Pixels don't receive light, photosites do.

Only if the sensors have the same number of pixels, but the number of pixels is irrelevant to the discussion, and it doesn't matter whether you call them 'pixels' or 'photosites' - it doesn't make any difference.

Even given this error and assuming he means photosites he makes the assumption that each photosite is four times larger in a FF sensor than in a 4/3 sensor. This ignores gaps that may exist between photosites and how these gaps may vary from one sensor to another.

That is a whole new and similarly fallacious argument. There isn't any systematic difference in the photon collecting efficiency of FF and FT pixels.

I also am the one that pointed out that the intensity of light per unit area is the same.

No, you aren't - it's been a part of the discussion from the start.

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